BST small element of the first K
Chinese
English
To a binary search tree, write a KthSmallest function to find the first of which K smaller elements.
Example
Example 1:
输入：{1,#,2},2
输出：2
解释：
1
\
2
第二小的元素是2。
Sample 2:
输入：{2,1,3},1
输出：1
解释：
2
/ \
1 3
第一小的元素是1。
Challenge
If the tree is modified often BST (insertion / deletion) and you need to very quickly find the K smaller elements of it? How would you optimize this KthSmallest function?
Notice
You may assume that k always effective 1 ≤ k ≤ 树的总节点数.
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: the given BST
@param k: the given k
@return: the kth smallest element in BST
"""
"""
nth = 0
result = None
def kthSmallest(self, root, k):
# write your code here
self.dfs(root, k)
return self.result
def dfs(self, root, k):
if not root:
return
self.dfs(root.left, k)
self.nth += 1
if self.nth == k:
self.result = root.val
self.dfs(root.right, k)
"""
""" template:
TreeNode pNode = root;
while (!s.isEmpty() || pNode != null) {
if (pNode != null) {
s.push(pNode);
pNode = pNode.left;
} else {
pNode = s.pop();
result.add(pNode.val);
pNode = pNode.right;
}
}
"""
def kthSmallest(self, root, k):
if not root:
return None
q = []
node = root
nth = 0
while q or node:
if node:
q.append(node)
node = node.left
else:
node = q.pop()
nth += 1
if nth == k:
return node.val
node = node.right
return None